All categories

3 years ago

Does anyone know what 5796 divided by 25 is?

Mathematics

2 answers:

anygoal [31]3 years ago

5 0

I think the answer is 231.84

defon3 years ago

5 0

Step-by-step explanation:

if we 5796 it will give us decimal number

this is the answer I have 231.84

You might be interested in

2/3 times what equals 10?

Aleks [24]

You have to divide 10 by $\frac{2}{3}$

$\frac{10}{1}$ ÷ $\frac{2}{3}$ ⇒ $\frac{10}{1}$ x $\frac{3}{2}$

You use a method called "flip-flop and multiply" where you flip the second fraction to multiply
$\frac{10}{1}$ × $\frac{3}{2}$ ⇒ $\frac{30}{2}$

Now divide the numerator (top) by the denominator (bottom)

$30
$ ÷ $3=15$

Now lets check:

$\frac{2}{3} * \frac{15}{1} = \frac{30}{3}$

$30$ ÷ $3=10

$

Answer: $\frac{2}{3}$ × $15=10$

8 0

3 years ago

Read 2 more answers

You finally get an allowance! You put $2 away. In January, $4 away. In February, $8 away. In March, $16 away. In April and follo

MrRa [10]

Add all the numbers together to find your answer.

___________________________________________

Remember : In April and followed this savings pattern through to December.

4 0

3 years ago

Pi High School ordered 40 science books. The next week, the school ordered 30 algebra books. The bill for the first order was $3

maria [59]

Answer:

Each algebra book costs $60.

Step-by-step explanation:

A=bill for algebra books; S=bill for science books=A+$360

S+A=$3960 Substitute for S

A+$360+A=$3960 Subtract $360 from each side.

2A=$3600 Divide each side by 2.

A=$1800 The order for 30 algebra books cost $1800.

Individual book=A/30

individual book=$1800/30=$60 ANSWER: Each algebra book cost $60.

4 0

3 years ago

According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 15x11 – 6x8 + x3 – 4x + 3?

scoundrel [369]

we have

$f(x) = 15x^{11} -6x^{8} + x^{3} - 4x + 3$

we know that

<u>The Rational Root Theorem</u> states that when a root 'x' is written as a fraction in lowest terms

$x=\frac{p}{q}$

p is an integer factor of the constant term, and q is an integer factor of the coefficient of the first monomial.

in this problem

the constant term is equal to $3$

and the first monomial is equal to $15x^{11}$ -----> coefficient is $15$

possible values of p are $1, and\ 3$

possible values of q are $1, 3, 5, and\ 15$

therefore

<u>the answer is</u>

The all potential rational roots of f(x) are

(+/-) $\frac{1}{15}$ ,(+/-) $\frac{1}{5}$ ,(+/-) $\frac{1}{3}$ ,(+/-) $\frac{3}{5}$ ,(+/-) $1$ ,(+/-) $3$

3 0

3 years ago

Read 2 more answers

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

3 years ago